The document NCERT Solutions(Part - 2) - Practical Geometry Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.

All you need of Class 6 at this link: Class 6

**Exercise 14.4****Question 1: Draw any line segment ****. Mark any point M on it. Through M, draw a perpendicular to ****. (Use ruler and compasses)****Answer 1:****Steps of construction:****(i) **Draw a line segment** **and mark a point M on it.**(ii) **Taking M as centre and a convenient radius, construct an arc intersecting the line segment at points C and D respectively.**(iii)** By taking centres as C and D and radius greater than CM, construct two arcs such that they intersect each other at point E.**(iv)** Join EM. Now is perpendicular to **Question 2:**** Draw any line segment . Take any point R not on it. Through R, draw a perpendicular to . (Use ruler and set-square)****Answer 2:****(i) **Draw a given line segment and mark a point R outside the line segment **(ii) **Place a set square on such that one of its right angles arm aligns along **(iii) **Now, place the ruler along the edge opposite to right angle of set square.**(iv)** Hold the ruler fixed. Slide the set square along the ruler such that the point R touches the other arm of set square.**(v)** Draw a line along this edge of set square which passes through point R. Now, it is the required line perpendicular to

**Question 3: ****Draw a line l and a point X on it. Through X, draw a line segment **** perpendicular to l. Now draw a perpendicular to **** to Y. (use ruler and compasses) ****Answer 3: ****Steps of construction:****(i) **Draw a line l and mark a point X on it.**(ii)** By taking X as centre and with a convenient radius, draw an arc intersecting the line l at points A and B respectively.**(iii)** With A and B as centres and a radius more than AX, construct two arcs such that they intersect each other at point Y.**(iv)** Join XY. Here is perpendicular to l

Similarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The line is perpendicular to at Y**Exercise 14.5****Question 1:**** Draw **** of length 7.3 cm and find its axis of symmetry.****Answer 1:** Axis of symmetry of line segment will be the perpendicular bisector of . So, draw the perpendicular bisector of .**Steps of construction:****(i)** Draw a line segment = 7.3 cm**(ii)** Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.**(iii)** Join CD. Then CD is the axis of symmetry of the line segment AB.**Question 2:****Draw a line segment of length 9.5 cm and construct its perpendicular bisector.****Answer 2:****Steps of construction:****(i)** Draw a line segment = 9.5 cm**(ii)** Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.**(iii)** Join CD. Then CD is the perpendicular bisector of .

**Question 3:**** Draw the perpendicular bisector of **** whose length is 10.3 cm.****(a) Take any point P on the bisector drawn.****Examine whether PX = PY.****(b) If M is the mid-point of ****, what can you say about the lengths MX and XY?****Answer 3:****Steps of construction:**

**(i)** Draw a line segment = 10.3 cm**(ii)** Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.**(iii)** Join CD. Then CD is the required perpendicular bisector of .

Now:

(a) Take any point P on the bisector drawn. With the help of divider we can check that

(b) If M is the mid-point of

**Question 4: ****Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.****Answer 4:****Steps of construction:****(i)** Draw a line segment AB = 12.8 cm**(ii)** Draw the perpendicular bisector of which cuts it at C. Thus, C is the midpoint of .**(iii)** Draw the perpendicular bisector of which cuts it at D. Thus D is the midpoint of.**(iv) **Again, draw the perpendicular bisector of which cuts it at E. Thus, E is the mid-point of .**(v)** Now, point C, D and E divide the line segment in the four equal parts.**(vi)** By actual measurement, we find that

**Question 5:****With **** of length 6.1 cm as diameter, draw a circle.****Answer 5:****Steps of construction:****(i)** Draw a line segment = 6.1 cm.**(ii)** Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of .**(iii)** Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment .

**Question 6:****Draw a circle with centre C and radius 3.4 cm. Draw any chord ****. Construct the perpendicular bisector **** and examine if it passes through C.****Answer 6:****Steps of construction:****(i) **Mark any point C on the sheet**(ii) **Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, turn compasses slowly to draw the circle. This is the required circle of 3.4 cm radius.

**(iii)** Mark any chord in the circle**(iv)** Now, taking A and B as centres, draw arcs on both sides of . Let these intersect each other at points D and E.**(v) **Join DE. Now DE is the perpendicular bisector of AB.

If is extended, it will pass through point C.**Question 7: Repeat Question 6, if happens to be a diameter.****Answer 7:****Steps of construction:****(i)** Mark any point C on the sheet.**(ii)** Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, Turn the compasses slowly to draw the circle. This is the required circle of 3.4 cm**(iii)** Now mark any diameter in the circle.**(iv)** Now taking A and B as centres, draw arcs on both sides of with radius more than . Let these intersect each other at points D and E.**(v)** Join DE, which is perpendicular bisector of AB.

Now, we may observe that is passing through the centre C of the circle.

**Question 8:****Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?****Answer 8:****Steps of construction:****(i) **Mark any point C on the sheet. Now adjust the compasses up to 4 cm and by placing the pointer of compasses at point C, turn the compasses slowly to draw the circle. This is the required circle of 4 cm radius**(ii) **Take any two chords in the circle**(iii)** By taking A and B as centres and radius more than half of draw arcs on both sides of AB. The arcs are intersecting each other at point E and F. Join EF which is perpendicular bisector of AB.**(iv)** Again take C and D as centres and radius more than half of draw arcs on both sides of CD such that they are intersecting each other at points G, H. Join GH which is perpendicular bisector of CD

We may observe that when EF and GH are extended they meet at the point O, which is the centre of circle**Question 9:****Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of **** and ****. Let them meet at P. Is PA = PB?****Answer 9:****Steps of construction:****(i) **Draw any angle with vertex as O.**(ii)** By taking O as centre and with convenient radius, draw arcs on both rays of this angle. Let these points are A and B**(iii)** Now take O and A as centres and with radius more than half of OA, draw arcs on both sides of OA. Let these intersects at points C and D respectively. Join CD**(iv)** Similarly we may find which is perpendicular bisector of . These perpendicular bisectors intersects each other at point P. Now measure PA and PB. They are equal in length.

**Exercise 14.6****Question 1:****Draw ∠ POQ of measure 75° and find its line of symmetry.****Answer 1:****Steps of construction:****(i) **Draw a line *l* and mark a point O on it.**(ii) **Place the pointer of the compasses at O and draw an arc of any radius which intersects the line *l* at A.**(iii)** Taking same radius, with centre A, cut the previous arc at B.**(iv)** Join OB, then ∠ BOA = 60 .**(v) **Taking same radius, with centre B, cut the previous arc at C.**(vi)** Draw bisector of ∠BOC. The angle is of 90 . Mark it at D. Thus, ∠DOA = 90^{o} **(vii)** Draw as bisector of ∠DOB. Thus, ∠POA = 75^{o.}

**Question 2: Draw an angle of measure 147 ^{o} and construct its bisector.**

Following steps are followed to construct an angle of measure 147

OC is the required bisector of 147

**Question 3:****Draw a right angle and construct its bisector.****Answer 3:****Steps of construction:****(i) **Draw a line PQ and take a point O on it.**(ii) **Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.**(iii)** Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.**(iv)** Join OC. Thus, ∠COQ is the required right angle.**(v)** Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.**(vi)** Join OD. Thus, is the required bisector of ∠COQ.

**Question 4:****Draw an angle of measure 153 ^{o} and divide it into four equal parts.**

Following steps are followed to construct an angle of measure 153

OF, OC, OE are the rays dividing 153

Thus, ∠BOA is required angle of 60° .

**(b)** **30**^{o}

**(i)** Draw a ray .**(ii)** Taking O as centre and convenient radius, mark an arc, which intersects at P.**(iii)** Taking P as centre and same radius, cut previous arc at Q.**(iv)** Join OQ. Thus, ∠BOA is required angle of 60^{o}.**(v)** Put the pointer on P and mark an arc.**(vi)** Put the pointer on Q and with same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30^{o}.

**(c) 90°****(i) **Draw a ray .**(ii)** Taking O as centre and convenient radius, mark an arc, which intersects at X.**(iii) **Taking X as centre and same radius, cut previous arc at Y.**(iv)** Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.**(v)** Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.**(vi)** Join OS and produce it to form a ray OB.

Thus, ∠BOA is required angle of 90^{0}.

**(d) 120°****(i) **Draw a ray .**(ii) **Taking O as centre and convenient radius, mark an arc, which intersects at P.**(iii) **Taking P as centre and same radius, cut previous arc at Q.**(iv)** Taking Q as centre and same radius cut the arc at S.**(v)** Join OS.

Thus, ∠AOD is required angle of 120°.

**(e) 45 ^{o} **

Thus, ∠MOA is required angle of 45°.

**(f) 135°**

**(i) **Draw a line PQ and take a point O on it.**(ii)** Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.**(iii) **Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.**(iv)** Join OR. Thus, ∠QOR = ∠POQ = 90°.**(v) **Draw the bisector of ∠POR.

Thus, ∠QOD is required angle of 135°.**Question 6:****Draw an angle of measure 45 ^{o} and bisect it.**

Following steps are followed to construct an angle of measure 45

OC is the required bisector of 45

Following steps are followed to construct an angle of measure 135

Let these intersect each other at C. Join OC.

OC is the required bisector of 135

Thus, ∠QMY = 40° and ∠PMY is supplementary of it.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

185 videos|229 docs|43 tests